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  #1  
Old July 20th, 2011, 10:03 PM
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Default Calculation needed...

I know there are some big-brain math folks out here...

So, you have a spaceship which burned at max. 2G acceleration for a period of 30 minutes, then maintaining that velocity...If we take-off in a ship capable of 5G acceleration, how long do we need to burn at 5G acceleration, before decelerating (or as my old Physics professor preferred, "applied negative acceleration")? And moreover, when would we catch-up or more precisely maintain an exact velocity with the other ship for the purpose of boarding?

I appreciate any and all responses!
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Old July 21st, 2011, 04:00 AM
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Quote:
Originally Posted by Fox's Master View Post
I know there are some big-brain math folks out here...

So, you have a spaceship which burned at max. 2G acceleration for a period of 30 minutes, then maintaining that velocity...If we take-off in a ship capable of 5G acceleration, how long do we need to burn at 5G acceleration, before decelerating (or as my old Physics professor preferred, "applied negative acceleration")? And moreover, when would we catch-up or more precisely maintain an exact velocity with the other ship for the purpose of boarding?

I appreciate any and all responses!
I may be a little rusty on this but it's late and I can't sleep.

frist assuming a starting velocity of 0 meters per second for both vehicals

after 30 minutes the 2G vehical (A) has a final velocity of 36,000 meters per second or 36 kilometers per second.

in those 30 minutes A has travelled 32,400 kilometers.

just to match velocities ship B with 5 g acceleration will take 720 seconds or 12 minutes.

in that time ship B will have travelled 12,960 Kilometers

so now both ships are travelling the same speed in the same direction but there still quite a distance betwen the two ships.

(32400-12960 = 19440) + (720*36 = 25,920) = 45360 kilometers seperation the two.

now fo our purposes to calculate time we just have to calculate how long it will take a ship accelerating at 5Gs to travelle 45360 KM with turn around at mid point, or at the 22,680KM mark.


22680000 M = 25T squared 0r 952.5 seconds to turn around point. or about 16 minutes to turn around.

so for ship A leaving with a 30 minute head start a 5g ship will take approximately 44 minutes from launch to catch and match velocites.

But I am willing to be corrected by some one else a little more current with thier physics.

R
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Last edited by Rover; July 21st, 2011 at 04:04 AM..
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Old July 21st, 2011, 06:15 AM
Icosahedron Icosahedron is offline
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Hmm, I'm a bit rusty on this, but I suspect there's a flaw in your calculation Rover in that you seem to assume both vehicles left at the same time, in which case there would be no chase...
...and you're doing it in two stages...
I dunno quite what's wrong, but I'm sure something is - maybe several somethings.

Looking at the OP, he hasn't specified how long the ship has been travelling at constant speed before the second ship gives chase. without that info, you can't solve the problem.

Even with that info, I think it's a simultaneous equation problem - but I could be wrong. Hell, just how rusty have I got?? Maybe I should work on this just for the exercise - I obviously need it!
Gimme the launch interval.
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Old July 21st, 2011, 08:32 AM
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The way understood the problem is:

ship A leaves the starting point with an initial velocity of zero. Accelerates for 30 minutes at 2g.(20 meters per second squared) then stops accelerating but continues to travel at a constant speed.

ship B leaves the same starting point with the same initial velocity of zero 30 minutes after A.

Ship B wants to match velocities and position with A so it can board.

I broke the problem down like this:

what is the velocity, and distance travelled by A in the first 30 minutes? 36 kilometres per second and 32,400 kilometres.

Vfinal = Vinitial + a * t

V = 0 + 20 * 1800

Vfinal = 36000 meters per second.

Vaverage + (Vinitial + Vfinal) over 2

V average = 18000 kilometres per second.

Distance travelled = Vaverage*time

distance = 1800 * 180

Distance = 3,240,000 meters or 32400 kilometres


how long does it take ship B, leaving 30 minutes after ship A to first match the velocity of ship A? 12 minutes and it has travelled 12,960 kilometres. Ship a has travelled an addition 720 seconds at 36 kilometres per second or 25,920 kilometres

Vfinal = Vinitial> + a * t

36,000 = 0 + 50 * t

solve for t = 720 seconds.


Both A and B are now travelling along the same line and at the same speed only 45,360 kilometres apart.

At this point Ship A has been travelling for 42 minutes and ship B 12. Ship A and B both stationary relative to each other at this instant, and the velocity of ship A does not change. For calculation purposes we can now treat ship A as if it were stationary. We now have to calculate how long it will take ship B to get to same relative position as A. It accelerates at full thrust 5g for half the relative distance then turns around and decelerates (negative acceleration) so it will be at the same

Distance = 0.5*Acceleration * (time squared)

22,680,000 meters = (0.5*50 meters per second squared) * (time squared)

22,680,000 = 25 * time squared

907,200 = time squared

time = square root of 907,200

time = 952.5 seconds. (rounded to 1 place)

so it will take and additional (2 * 952.5 seconds) for ship B to match position and velocity of A.

That is 1905 seconds or 31.75 minutes (I rounded to 32)

So from the point ship B leaves it will take approximately 44 minutes (12+32)

Ship A will have been running for 74 minutes
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Last edited by Rover; July 21st, 2011 at 08:37 AM..
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Old July 22nd, 2011, 01:53 AM
Icosahedron Icosahedron is offline
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Yep. Now I've sat down and thought, instead of skim-reading it, you are correct - IF your initial assumption holds that ship B starts out at the moment ship A stops accelerating.

That split not only seems valid, it's easier. You are obviously less rusty than me.

Over to Fox's Master to confirm (or otherwise) the assumption.
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Old July 22nd, 2011, 05:24 AM
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Rover and Icosahedron,

Thank you ~ I was not very clear with some of the points. Yes, the second ship gave chase approximately 30 minutes after the first ship passed the space station. Your elaboration of the problem is exquisite...thank you very much!
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Old July 22nd, 2011, 05:47 AM
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It has been a while since I calculated anything like that. It was kind of fun. I don't think i could hav calculated an interception point if Ship A ha been under constant Acceleration.

Thanks and you're welcome.

R
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Old July 22nd, 2011, 11:46 AM
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Thank you very much for the original post and also for the very entertaining replies. I've not had to, as many of us I suspect, use that part of my brain in a very long time.

Of course instead of giving chase you could have simply press the GM remote destruct button and then let the fools roll up new characters.
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Old July 23rd, 2011, 12:16 AM
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One change to the initial assumption...the ship that we're trying to intercept is 1 million km from the station which we start our chase.
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Old July 23rd, 2011, 01:52 AM
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Quote:
Originally Posted by Fox's Master View Post
One change to the initial assumption...the ship that we're trying to intercept is 1 million km from the station which we start our chase.
So A and B are not starting from the same place?
Talking without engaging my brain properly again, I'd say that adds to the distance B has to cover during the 'A is relatively stationary' phase. Have a go at figuring that yourself using Rover's example.

Edit: Whoa!... I just looked upthread - you say Ahip A is passing a space station. Do you mean that Ship A's trajectory passes Ship B's launch point at a distance of 1Mkm, so that B has to travel at an angle to A's trajectory in order to catch up? (A is headed 'North' but is 1Mkm 'East' of B's launch point?) That's a whole different ball-game. I'm not sure I even want to imagine that one. If I scrape that much rust off my brain there may be nothing left! Any professional Astrogators here? Anyone work for NASA (and not as a janitor)?

Last edited by Icosahedron; July 23rd, 2011 at 02:05 AM..
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